Answer:
E = 2.7 x 10¹⁶ J
Explanation:
The release of energy associated with the mass can be calculated by Einstein's mass-energy relation, as follows:

where,
E = Energy Released = ?
m = mass of material reduced = 0.3 kg
c = speed of light = 3 x 10⁸ m/s
Therefore,

<u>E = 2.7 x 10¹⁶ J</u>
Answer:
This slide shows the three forces that act on a baseball in flight. The forces are the weight, drag, and lift. Lift and drag are actually two components of a single aerodynamic force acting on the ball. Drag acts in a direction opposite to the motion, and lift acts perpendicular to the motion
The right answer for the question that is being asked and shown above is that: "A.tectonic activity concentrated in certain areas." A piece of evidence did Alfred Wegener use to develop the theory of continental drift is that <span>A.tectonic activity concentrated in certain areas</span>
Answer:
Explanation:
Given that:
mass of stone (M) = 0.100 kg
mass of bullet (m) = 2.50 g = 2.5 ×10 ⁻³ kg
initial velocity of stone (
) = 0 m/s
Initial velocity of bullet (
) = (500 m/s)i
Speed of the bullet after collision (
) = (300 m/s) j
Suppose we represent
to be the velocity of the stone after the truck, then:
From linear momentum, the law of conservation can be applied which is expressed as:





∴
The magnitude now is:


Using the tangent of an angle to determine the direction of the velocity after the struck;
Let θ represent the direction:

